The Distributive Property for 7th Grade: Using additional geometric shapes

Let us begin by reminding ourselves of the formula to calculate the area of a rectangle: width X length

$S=w⋅h$

When:

S = area

w = width

h = height

We take data from the sides of the rectangle in the figure.$w=b+8$$h=a+3$

We then substitute the above data into the formula in order to calculate the area of the rectangle:

$S=w⋅h = (b+8)(a+3)$

We use the formula of the extended distributive property:

$(a+b)(c+d)=ac+ad+bc+bd$

We substitute once more and solve the problem as follows:

$S=(b+8)(a+3)=(b)(a)+(b)(3)+(8)(a)+(8)(3)$

$(b)(a)+(b)(3)+(8)(a)+(8)(3)=ab+3b+8a+24$

Therefore, the correct answer is option B: ab+8a+3b+24.

Keep in mind that, since there are only addition operations, the order of the terms in the expression can be changed and, therefore,

$ab+3b+8a+24=ab+8a+3b+24$

Let us begin by reminding ourselves of the formula to calculate the area of a rectangle: width X height

$S=w⋅h$

Where:

S = area

w = width

h = height

We must first extract the data from the sides of the rectangle shown in the figure.

$w=3y$$h=y+3z$

We then insert the known data into the formula in order to calculate the area of the rectangle:

$S=w⋅h=(y+3z)(3y)$

We use the distributive property formula:

$a\left(b+c\right)=ab+ac$

We substitute all known data and solve as follows:

$S=(y+3z)(3y)=(3y)(y+3z)$

$(3y)(y+3z)=(3y)(y)+(3y)(3z)$

$(3y)(y)+(3y)(3z)=3y^2+9yz$

Keep in mind that because there is a multiplication operation, the order of the terms in the expression can be changed, hence:

$(y+3z)(3y)=(3y)(y+3z)$

Therefore, the correct answer is option D: $3y^2+9yz$

Let's begin by reminding ourselves of the formula to calculate the area of a rectangle: width X length

$S=w⋅h$

Where:

S = area

w = width

h = height

We extract the data from the sides of the rectangle in the figure.

$w=x+5$$h=y+2$

We then substitute the above data into the formula in order to calculate the area of the rectangle:

$S=w⋅h=(x+5)(y+2)$

We use the formula of the extended distributive property:

$(a+b)(c+d)=ac+ad+bc+bd$

We once again substitute and solve the problem as follows:

$S=(x+5)(y+2)=(x)(y)+(x)(2)+(5)(y)+(5)(2)$

$(x)(y)+(x)(2)+(5)(y)+(5)(2)=xy+2x+5y+10$

Therefore, the correct answer is option C: xy+2x+5y+10.